Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are: and . The free ends of rods A and B maintained 400 K, 200 K, respectively. The temperature

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Solution

Q.
Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are: $\frac{L_{A}}{L_{B}} = \frac{1}{2}, \frac{r_{A}}{r_{B}} = 2$ and $\frac{K_{A}}{K_{B}} = \frac{1}{2}$ . The free ends of rods A and B are maintained at 400 K, 200 K, respectively. The temperature of rods interface is ________ K, when equilibrium is established. [2025]

Ans.
(360)

$R_{1} = \frac{l_{1}}{K_{1} A_{1}}, R_{2} = \frac{l_{2}}{K_{2} A_{2}}$

$\frac{d Q}{d t} = \frac{△ T}{R}$

${(\frac{d Q}{d t})}_{1} = {(\frac{d Q}{d t})}_{2}$

$\frac{400 - T}{R_{1}} = \frac{T - 200}{R_{2}}$

$\frac{400 - T}{T - 200} = \frac{R_{1}}{R_{2}} = (\frac{l_{1}}{l_{2}}) {(\frac{r_{2}}{r_{1}})}^{2} \times \frac{K_{2}}{K_{1}}$

$\frac{400 - T}{T - 200} = \frac{1}{2} \times {(\frac{1}{2})}^{2} \times 2 = (\frac{1}{4})$

1600 – 4T = T –200

5T = 1800 $\Rightarrow$ T = 360 K

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